Find the equation to the tangent to the curve x=cos(2y+pi) at (0, pi/4)

Normally to find a tangent we want to work out dy/dx, but since this equation is x=something, it's much easier to work out dx/dy first, then we get dy/dx by doing 1/(dx/dy)=dy/dx.
By the chain rule, dx/dy = -sin(2y+pi)2, since cos differentiates to -sin, and we need to remember to differentiate the bit in the brackets too, which is why we multiply by 2. Now let's substitute in our x and y values, and we get that dx/dy = -2sin(3pi/2) and (0, pi/4), which equals 2. So by using the little formula I gave earlier, we get dy/dx=1/2 here. So we know the tangent line has gradient 1/2, and passes through the point (0, pi/4), so we use the equation y=mx+c with m=1/2, which gives us c=pi/4, and the equation of the tangent line is y=1/2x + pi/4.

SJ
Answered by Sarah J. Maths tutor

10416 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given that y = 5x(3) + 7x + 3, find A) dy/dx B) d2y/dx2


How exactly does integration by parts work?


Let f(x) and g(x) be two odd functions defined for all real values of x. Given that s(x)=f(x)+g(x), prove that s(x) is also an odd function.


How do I integrate cos^2x with respect to x?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences